Proof By Induction - Divisibility by $7$. I am attempting:

My solution is: But I am not sure where I am going wrong. The answer I get is not divisible by 7.

 A: HINT:
You have to prove the truth of $p(k+1)$ using $p(k)$, so you have to take out something from $p(k$) and then apply it to $p(k+1)$ to establish its truth.
As you have assumed that $p(k)$ is true. So, $4^{k+1}+5^{2k-1}$ must be divisible by 7 say it is $7m$ where $m$ is an integer. So you get $4^{k+1}+5^{2k-1}=7m$. Some flipping will give you $5^{2k-1}=7m-4^{k+1}$. Now How does $p(k+1)$ looks like??
It will look like $4^{k+2}+5^{2k+1}$. If we prove that $4^{k+2}+5^{2k+1}$ is divisible by $7$ then we are done. Try using  $5^{2k-1}=7m-4^{k+1}$ to proceed further.
A: Hint: $4^{n+2}+5^{2n+1}=4(4^{n+1})+25(5^{2n-1})=4(4^{n+1}+5^{2n-1})+21(5^{2n-1})$.
A: First, show that this is true for $n=1$:
$4^{1+1}+5^{2-1}=21$
Second, assume that this is true for $n$:
$4^{n+1}+5^{2n-1}=7k$
Third, prove that this is true for $n+1$:
$4^{n+2}+5^{2n+1}=$
$4(4^{n+1})+25(5^{2n-1})=$
$(25-21)(4^{n+1})+25(5^{2n-1})=$
$25(4^{n+1})-21(4^{n+1})+25(5^{2n-1})=$
$25(4^{n+1})+25(5^{2n-1})-21(4^{n+1})=$
$25(\color\red{4^{n+1}+5^{2n-1}})-21(4^{n+1})=$
$25(\color\red{7k})-21(4^{n+1})=$
$7(25k)-21(4^{n+1})=$
$7(25k-3(4^{n+1}))$

Please note that the assumption is used only in the part marked red.
A: Hint
Write like this:
$$4^{k+2}-4^{k+1}+5^{2k+1}-5^{2k-1}=3\cdot4^{k+1}+24\cdot5^{2k-1}=3\cdot(4^{k+1}+5^{2k-1})+21.5^{2k-1}$$
Can you finish? 
P.S: Use hyp induction for $k$ and note that $7|21$.
P.S: You can also prove that without use induction.
A: Better consider the expression: $$\delta(k+1) = 4^{k+2} + 5^{2k+1} = 4\times 4^{k+1} + 5^2\times 5^{2k-1} = 4[7m-5^{2k-1}] +5^{2k-1}(5^2) = [5^{2k-1} \times (5^2-4) +4(7m)] \mid 7$$ We considered $\delta(k) = 4^{k+1} +5^{2k-1} = 7m$ , since we know $7\mid \delta(k)$. Hope it helps.
A: Note $5^2\equiv 4$ and $5^{-1}\equiv 3\mod 7$. Using this, we have:
$$4^{n+1}+5^{2n-1}\equiv 4\cdot 4^n+(5^2)^n\cdot5^{-1}\equiv4\cdot 4^n+3\cdot 4^n=7\cdot4^n\equiv 0\mod 7.$$
A: Before considering your proof, let's gather some insight from a simpler proof using congruences. Below the inductive step follows very simply by using $\,\rm{\color{#C00}{CPR}} = $ Congruence Product Rule to multiply the first two congruences 
$$\begin{align} 
{\rm mod}\,\ 7\!:\qquad\  4\,\ &\equiv\,\  5^{\large 2}\\[0.3em]
4^{\large K+1}&\equiv -5^{\large 2K-1}\ \ \ {\rm i.e.}\ \ P(K)\\
\overset{\rm{\color{#C00}{CPR}}}\Longrightarrow\ \ \ 4^{\large K+2}&\equiv -5^{\large 2K+1}\ \ \ {\rm i.e.}\ \ P(K\!+\!1)
\end{align}$$

The common inductive proofs using divisibility in other answers effectively do the same thing, i.e. they repeat the proof of the Congruence Product Rule in this special case, but expressed in divisibility vs.  congruence language (e.g. see here).  But the product rule is much less arithmetically intuitive when expressed as unstructured divisibilities, which greatly complicates the discovery of the inductive step. I explain this at length in other answers, e.g. see here.
If congruences  are unfamiliar then you can instead use the rule in divisibility form as below. This will allow you to structure the induction in the above intuitive arithmetical Product Rule form.  
$$\begin{align} {\rm mod}\,\ m\!:\, A\equiv a,\, B\equiv b&\ \ \,\Longrightarrow\,\ \ AB\equiv ab\qquad\text{Congruence Product Rule}\\[3pt]
m\mid A-a,\ B-b&\,\Rightarrow\, m\mid AB-ab\qquad\text{Divisibility Product Rule}\\[4pt] 
{\bf Proof}\quad (A-a)B+a(B&-b)\, = AB-ab\end{align}$$

To finish the proof that you started we can proceed as follows
$$\begin{align} f(k\!+\!1) - f(k) &=\, 3 \cdot 4^{\large k+1}+ \color{#0a0}{24}\cdot  5^{2k-1}\\
&=\, 3 \cdot 4^{\large k+1}+ \color{#0a0}3 \cdot  5^{2k-1} + \color{#0a0}{21}\cdot  5^{2k-1}\\
&=\, 3\, f(k) + 7n\\
\Rightarrow\qquad f(k\!+\!1)\, &=\, 4\, f(k) + 7n\\[0,3em]
\Rightarrow\ \ 7\mid f(k\!+\!1)\,\ &{\rm if}\,\  7\mid f(k), \ \ {\rm i.e.}\ \ P(k\!+\!1)\ \ {\rm if}\ \ P(k)
\end{align}$$
Note that the above says that $\ f(k\!+\!1)\equiv 4\,f(k)\ \pmod{7}\,$
so an easy induction shows that $\ f(k)\equiv 4^{\large k-1}\, f(1)\pmod 7,\ $ so $\ 7\mid f(k)\iff 7\mid f(1)$ 
Now how using the Product Rule as above makes it much clearer that incrementing the index amounts simply to multiplication by $\,4,\,$ when viewed modulo $\,7.\,$ Once that innate arithmetical structure hs been revealed, the proof is easy.
